You visited us 52 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Displacement Current

A parallel pl...

Question

A parallel plate capacitor has plate area $100 c m^{2}$ and separation between the plates is 1 cm. A glass plat $(k_{1} = 6)$ of thickness 6 mm and an ebonite plate $(k_{2} = 4)$ of thickness 4 mm are inserted. Find C:

4.085 pF

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40.85 pF

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

.4085 nF

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40.85 nF

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 40.85 pF
Capacitors are in series so equivalent capacitance is given by:
$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{ϵ_{0} A k_{1} k_{2}}{k_{1} d_{2} + k_{2} d_{1}} = \frac{8.85 \times 10^{- 12} \times 10^{- 2} \times 6 \times 4}{(6 \times 6 + 4 \times 4) \times 10^{- 3}} = 4.085 \times 10^{- 11} F$

Suggest Corrections

Similar questions

Q. A parallel-plate capacitor has plate area 100 cm² and plate separation 1⋅0 cm. A glass plate (dielectric constant 6⋅0) of thickness 6⋅0 mm and an ebonite plate (dielectric constant 4⋅0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.

Q. A parallel plate capacitor with air between the plates has a capacitance of 9

p F

. The separation between its plates is

d

. The space between the plates is now filled with two dielectrics. One of the dielectric has dielectric constant

K_{1} = 3

and thickness

\frac{d}{3}

while the other one has dielectric constant

K_{2} = 6

and thickness

\frac{2 d}{3}

. Capacitance of the capacitor is now :

The plates of a parallel plate capacitor are charged to $200 V$ and then, the charging battery is disconnected. Now, a dielectric slab of dielectric constant $5$ and thickness $4 m m$ is inserted between the capacitor plates. To maintain the original capacity, the increase in the separation between the plates of the capacitor is:

Q. A parallel plate capacitor has area

6 {cm}^{2}

and separation

3 mm

. The gap between the plates is filled with three dielectric materials of equal thickness as shown. The equivalent dielectric constant of a material which would give the same capacitance when fully inserted in the capacitor, would be

Q. The area of each plate of a parallel plate capacitor is

2 m^{2}

. The space between the plates is filled with materials of dielectric constants 2, 3, and 6 and their thickness are 0.4 mm,0.6 mm and 1.2 mm respectively. The capacitance of the capacitor will be :

Join BYJU'S Learning Program

A parallel plate capacitor has plate area 100cm2 and separation between the plates is 1 cm. A glass plat (k1=6) of thickness 6 mm and an ebonite plate (k2=4) of thickness 4 mm are inserted. Find C:

The correct option is B 40.85 pFCapacitors are in series so equivalent capacitance is given by: Ceq=C1C2C1+C2=ϵ0Ak1k2k1d2+k2d1=8.85×10−12×10−2×6×4(6×6+4×4)×10−3=4.085×10−11F

A parallel plate capacitor has plate area $100 c m^{2}$ and separation between the plates is 1 cm. A glass plat $(k_{1} = 6)$ of thickness 6 mm and an ebonite plate $(k_{2} = 4)$ of thickness 4 mm are inserted. Find C: